It can be shown(eg by Ceva trigonometric form) easily that $H$ lies on the altitude $AA'$( $A'\in{BC}$) from $A$ to $BC$. Then since $A'HGD$ and $AFDE$ are cyclic quadrilaterals : $BH\cdot{BG}=BA'\cdot{BD}=BF\cdot{BA}$ hence $AFHG$ is cyclic. In a similar way: $CI\cdot{CH}=CE\cdot{CA}=CD\cdot{CA'}$ and hence $AEIH$ is cyclic, too.

Sailor wrote: $H$ lies on the altitude $AA'$( $A'\in{BC}$) from $A$ to $BC$. That is the key to this problem. Many contestants failed to notice this and thus did not get this problem right. It can be proved by trigonometry, pure geometry, or even using Cartesian coordinates!